A light bulb with a resistance R = 75 Ohm was connected to a current source with an internal resistance r = 5 Ohm

A light bulb with a resistance R = 75 Ohm was connected to a current source with an internal resistance r = 5 Ohm. If the amount of heat equal to Q = 1125 J is released to the light bulb in 4 minutes, then the EMF of the source is equal to.

Power P dissipated per bulb:

P = Q / t,

where Q is the amount of heat,

t is time.

The same power P, expressed in terms of resistance R and current through the bulb I:

P = I2R.

We find the current through the light bulb I:

Q / t = I2R;

I2 = Q / tR;

I = √Q / tR.

Ohm’s law for a complete circuit:

I = E / (R + r),

where E – EMF,

r – internal resistance.

E = I * (R + r) = √ (Q / tR) * (R + r) =

= √ (1125 J / (240 s * 75 Ohm)) * (75 Ohm + 5 Ohm) =

= √ (45 / (240 * 3)) * 80V = √1 / 16 * 80 = 20V.

Answer: 20 V.